In the $A C$ circuit shown in the figure, all the voltmeters and the ammeter are ideal and can directly measure rms values. Reading of the ammeter (A) is $2 A .$ The voltmeters $V_{1}, V_{3} a n d V_{4}$ read $100 V, 120 V a n d 140 V$ respectively.

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The readings of the voltmeters $V_{2}$ and $V_{5}$ are nearly $98 V$ and $22 V$ respectively.

The source voltage is nearly $102.4 V$

The resistance $R = 50 Ω$

Reactance of the circuit is nearly $11 Ω$

answer is A, B, C, D.

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Detailed Solution

(a) $V_{1}^{2} + V_{2}^{2} = V_{4}^{2}$ or, $V_{2}^{2} = {(140)}^{2} - {(100)}^{2} \Rightarrow V_{2} = 98 V$
The voltage across L and C are at 180° phase difference.
$∴ V_{5} = V_{3} - V_{2} = 120 - 98 = 22 V$
(b) $V = \sqrt{100^{2} + 22^{2}} = 102.4 V$
(c) $I R = V_{1}$
or, $R = \frac{100}{2} = 50 Ω$
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